Question: $\dfrac{d}{dx}(4x^4-7x^3+3)=$
Explanation: According to the sum rule, the derivative of $4x^4-7x^3+3$ is the sum of the derivatives of $4x^4$, $-7x^3$, and $3$. The derivatives of these terms can be found using the power rule : $\dfrac{d}{dx}(x^n)=n\cdot x^{n-1}$ For example, this is the derivative of the first term: $\begin{aligned}\dfrac{d}{dx}(4x^4)&=4\dfrac{d}{dx}(x^4)&&\gray{\text{Constant multiple rule}}\\\\ &=4\cdot (4x^3)&&\gray{\text{Power rule}}\\ \\ &=16x^3\end{aligned}$ Here is the complete differentiation process: $\begin{aligned} &\phantom{=}\dfrac{d}{dx}(4x^4-7x^3+3) \\\\ &=4\dfrac{d}{dx}(x^4)-7\dfrac{d}{dx}(x^3)+\dfrac{d}{dx}(3)&&\gray{\text{Basic differentiation rules}} \\\\ &=4\dfrac{d}{dx}(x^4)-7\dfrac{d}{dx}(x^3)+0&&\gray{\text{Constant rule}} \\\\ &=4\cdot 4x^3-7\cdot3x^2&&\gray{\text{The power rule}} \\\\ &=16x^3-21x^2 \end{aligned}$ In conclusion, $\dfrac{d}{dx}(4x^4-7x^3+3)=16x^3-21x^2$.